Simple Definite Integral 2

Calculus Level 4

$\large{ I = \int_1^{16} \dfrac{\ln(x)}{\sqrt{x}(x+4)} \, {d}x}$

If $I$ can be expressed as $A\ln(A) \tan^{-1} \left( \dfrac{B}{C} \right)$ where $A,B,C$ are positive integers with $\gcd(B,C) = 1$ and $A$ is a prime number. Then find the value of $A+B+C$ .

The answer is 9.

2 solutions

Rajdeep Dhingra
Dec 9, 2015

$\text{We have I }= \int_{1}^{16}{\dfrac{\ln(x)}{\sqrt{x}(x + 4)} dx} \\ \text{Make a substitution } t = \sqrt{x} \\ \text{We get } I = 2 \int_{1}^{4}{\dfrac{2 \ln(t)}{t^2 + 4} dx} \\ \text{Make a substitution } 2\tan(y) = t \\ \text{We get } I = 2\int_{\arctan(1/2)}^{\arctan(2)} {\ln(2\tan(y)) dy} \quad \boxed{_1}$ Now we will use properties of Definite Integrals (Property 3 from Basic Properties of Integrals )

We get

$I = 2\int_{\arctan(1/2)}^{\arctan(2)} {\ln(2\cot(y)) dy}\quad \boxed{_2} \\ \text{Adding 1 and 2 we get } \\ I = \int_{\arctan(1/2)}^{\arctan(2)}{\ln(4)dy}\\ I = 2\ln(2) \arctan(3/4)$

Moderator note:

Good approach of using multiple change of variables to simplify the integral further. How does one come up with $2 \tan y = t$ as the substitution to remove the denominator?

beautiful solution

Prabhat Koutha - 4 years, 10 months ago

Thanks. $\ddot\smile$

Rajdeep Dhingra - 4 years, 10 months ago

I think you meant $\ln(2\cot(y))$ in the first line of the second paragraph.

Kenny Lau - 5 years, 6 months ago

Ohh yes ! Thanks

Rajdeep Dhingra - 5 years, 6 months ago

Response to Challenge Master note : When ever we have $x^2 + a^2$ in the denominator, the best approach is by putting $x = a \tan(t)$ .

Rajdeep Dhingra - 5 years, 6 months ago

how does a baby learn his first steps or how does one realise that it is with our mouth that we must eat. literally it is the first substitution they make you learn in calculus class.

Antara Chatterjee - 3 years, 1 month ago

Prakhar Bindal
Nov 22, 2016

Substitute x = 4tan^2(y)

Simple Definite Integral 2

The answer is 9.

2 solutions

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